How do you factor #y^2-16+64#?

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Answer 1 · 2016-01-24T21:57:13

This trinomial is a perfect square trinomial, so can be factored as #(a-b)^2#

The square root of 64 is 8 and the square root of #y^2# is y.

The middle term should be of the form 2ab. In this case, a=y and b= 8

So, 2(y)(8), or 16y, which is the middle term. This short proof justifies that it is indeed a perfect square trinomial.

#y^2# - 16y + 64
= (y - 8)(y - 8)
= #(y - 8)^2#

Your answer is #(y -8)^2#

Exercises:

a). #y^2# - 16

B) #y^2# + 8y + 16

C) #y^2# + 16

D) #y^2# - 8y + 16

#4x^2# + 28x + 49

Hopefully you understand now!